All Questions
Tagged with ordinals axiom-of-choice
46
questions
4
votes
0
answers
105
views
Why does proof of Zorn's lemma need to use the fact about ordinals being too large to be a set?
I'm not understanding why its necessary to invoke the knowledge about ordinals in order to prove Zorn's lemma.
The Hypothesis in Zorn's lemma is
Every chain in the set Z has an upper bound in Z
Then ...
- 41
2
votes
1
answer
102
views
Using a form of the Axiom of Choice to prove the Well Ordering Theorem
I have been studying ordinals on my own for a few weeks now, and I have just gone through the proof for Hartogs' lemma, that given $S$ a set, $\exists \alpha$ an ordinal s.t. there is no injective ...
- 93
2
votes
1
answer
104
views
Can $ZF$ construct this function?
Is it possible in the $ZF$ theory to construct a function with domain $\omega_1$ (the set of all countable ordinals) that maps each countable ordinal $\alpha$ to a bijection between $\alpha$ and $\...
- 555
0
votes
1
answer
32
views
Reordering a Sequence of Sets Whose Union is the Whole Set
I have a set $ B $ that can be written as $ B = \cup_{\nu < \lambda} B_{\nu} $ where $ \kappa $ is the cardinality of $ B $, that is uncountable, and $ \vert B_{\nu} \vert < \kappa $ with $ \...
- 51
1
vote
0
answers
166
views
Ordinal embedding in $\mathbb{R}$, and the use of $\mathsf{AC}$
Prove that if an ordinal $\alpha$ can be embedded (in an order preserving manner) into $\Bbb{R}$, then $\alpha$ is countable.
I managed to prove this using $\mathsf{AC}$ (countable choice is ...
- 1,371
2
votes
1
answer
111
views
Infinite Lexicographic Order on Bijections is Well-Order
Problem. Prove, without using $\mathsf{AC}$ if possible , that if $\alpha$ and $\beta$ are ordinals such that $\alpha$ is countable and $\beta>1$, then $\alpha^\beta$ is countable.
The induction ...
- 1,371
2
votes
1
answer
102
views
Finding a winning strategy for a game in ZF + “$\omega_1$ is regular”.
Say two players ($A$ and $B$) are playing a game of length $\omega$ on $\omega_1$ in which they play a countable ordinal turn by turn (starting with $A$) and form a sequence of countable ordinals, $(...
- 1,316
2
votes
1
answer
152
views
Use of the axiom of choice in a first course in ordinals
(It's likely that this entire question is answered with a "yes, you're right" comment. I'm just proceeding with extreme caution after discovering an error in my notes.)
I was reading my ...
- 36.4k
2
votes
1
answer
180
views
Why does global choice imply a bijection between all proper classes?
According to this old answer, global choice implies its own version of the well-ordering theorem. Namely that
Every proper class can be put in bijection with the class of ordinals.
There are two ...
- 1,602
0
votes
1
answer
40
views
Finding surjections from some subsets of $\omega^2$ to $\omega$ many at most countable ordinals without choice.
Set $A_k = \{\gamma \ | \ \omega \cdot k \leq \gamma < \omega \cdot (k+1) \}$, and let $B$ be a set of ordinals such that $B = \{ \alpha_n \ | \ n \in \omega \}$ where each $\alpha_n$ is an at ...
- 1,316
1
vote
1
answer
135
views
In $\mathsf{ZF}$, for each set $A$ exists an ordinal $\alpha$ such that no surjection $A\to \alpha$ exists
Some observations.
The statement can easily be proven in $\mathsf{ZFC}$, but we do not
assume that $\mathsf{AC}$ holds.
If one proves that $S=\{\beta \mid \text{there is a surjection $A\to
\beta$}...
- 4,743
1
vote
0
answers
28
views
Is this proposition regarding equipotence true, without the needing axiom of choice? [duplicate]
Proposition: let $A$ and $B$ be arbitrary sets. Then, there exists a set $C$ such that $A\cap C=\varnothing$ and $C\sim B$.
This proposition seems really easy at first, but when I tried to prove it, ...
- 464
-2
votes
1
answer
72
views
Can $\omega_2$ be the order type of a countable union of countable sets of ordinals?
Working in ZF(wihtout choice), according to this answer: $\omega_2$ is not the countable union of countable sets.
Can $\omega_2$ be the order type of a countable union of countable sets of ordinals?
- 4,631
0
votes
0
answers
30
views
Can ZF prove the cartesian product of an ordinal by itself to be subnumerous to that ordinal? [duplicate]
Does ZF (without choice) prove the following?
$\forall \ ordinal \ \alpha \ \exists f( f:\alpha \times \alpha \hookrightarrow \alpha)$
where: $ \alpha\times \alpha $ is the cartesian product of $\...
- 4,631
1
vote
1
answer
141
views
How does $\mathsf{ZF}$ prove the countability of ordinals coming out of functions such as the ordinal collapse function?
I am reading about fast growing hierarchies and the ordinal functions that are used in the indices, which allow us to categorise fast growing functions by selecting fundamental sequences for these ...
- 509